Solve $\sqrt{2x^2 - 1} = x$.

$$\sqrt{2x^2 - 1} = x$$ $$2x^2 - 1 = x^2 \tag{Squared both sides}$$ $$x^2 - 1 = 0$$ $$(x - 1)(x + 1) = 0$$

Wolfram alpha says that it's only true for $x=1$ (Link).

What's wrong in the solution above?


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    $\begingroup$ Wolfram alpha uses the standard convention that the square root of a positive real number means the positive of the two square roots of the positive real number. In this case there are two solutions to $2x^2 -1 = x^2$, but only one solution to $\sqrt{2x^2 - 1} = x$ because to satisfy the second equality $x$ must be nonnegative. Compare $\sqrt{x^2} = |x|$. $\endgroup$ – Dan Fox Apr 6 '17 at 7:41
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    $\begingroup$ This is a good example of why you should check your solutions. Try inserting $x=-1$ in the original equation and see what happens. $\endgroup$ – Arthur Apr 6 '17 at 7:44

The thing that is wrong is that not all steps are equivalences. In squaring both sides of an equation you only forms an implication. That is if $a=b$ then $a^2=b^2$ but the reverse does not hold.

The reason the squaring of the sides does not reverse is that the square of a number is the same as the square of it's negative. That is $(-a)^2 = a^2$.

All the other steps forms equivalencies.

This means that if $\sqrt{2x^2-1}=1$ we certeinly have that $x=1\lor x=-1$, but not the reverse (the second statement is true if $x=-1$, but the first isn't).


There's nothing wrong with the proof, you are just confused about what you proved.

What you proved is that if $x$ solves the equation $\sqrt{2x^2-1} = x$, then $x$ is either $1$ or $-1$.

What you did not prove is that if $x$ is either $1$ or $-1$, then $x$ solves the equation $\sqrt{2x^2-1} = x$.

The thing is that you made a "non-reversible" step in simplifying equations, i.e. you went from the equation $a=b$ to the equation $a^2=b^2$. This is a completely valid step, except it cannot be reversed. In other words, the logical statement $$a=b\implies a^2=b^2$$ is true, but the logical statement $$a^2=b^2\implies a=b$$ is not true, as the example $a=1, b=-1$ clearly shows.

Now, there's nothing wrong with making nonreversible steps when solving equations, the only thing is that the may result in an equation with more solutions than the original one. So, you won't lose any solutions, but you may generate some other quasi-solutions. Basically, in the end, you won't get solutions, you will get candidates.

So, when making such steps, you need to

  1. Notice when you make them (squaring the equation is a typical example)
  2. Check the solutions when you are done.

In your case, your two candidates for the solution are $x=1$ and $x=-1$, and you need to check them.

  1. Checking $x=1$: $\sqrt{2x^2-1} = \sqrt{2\cdot 1^2-1}=\sqrt{1}=1=x$, so $x=1$ is a solution.
  2. Checking $x=-1$: $\sqrt{2x^2-1} = \sqrt{2\cdot (-1)^2-1}=\sqrt{1}=1\neq -1 =x$ , so $x=-1$ is not a solution.
  • $\begingroup$ Can the downvoter please explain why this answer is bad so much that it deserves a downvote? $\endgroup$ – 5xum Apr 6 '17 at 7:56

The initial equation implies that $x$ is non-negative (by definition of the square root), hence you must reject the solution $x=-1$, which was introduced by the squaring.

  • $\begingroup$ I see, x can't be negative because it's under square root? $\endgroup$ – dud3 Apr 6 '17 at 8:04
  • $\begingroup$ @dud3: not at all. Try again... $\endgroup$ – Yves Daoust Apr 6 '17 at 8:11

$$\sqrt{f(x)}=g(x)\Leftrightarrow \begin{cases} \ f(x)=\left(g(x)\right)^2 \\ g(x)\ge0 \end{cases}$$

Then $$\sqrt{2x^2-1}=x\Leftrightarrow \begin{cases} \ 2x^2-1=x^2 \\ x\ge0 \end{cases}$$


If you want to solve this equation you are doing valid steps until the last one, but you should notice that the first step is not an "if and only if" equivalence and the right direction of the equivalence is the only true. So now if you solve the both factors equaling them to 0 and computing the solutions $x\in \{-1,1\}$ you will have the solutions for the last equation, not for the original equation! Now you have to check if they are valid in the original equation!

Caution, as we said before in the first step you are squaring $x$ so maybe one of the previous solutions isn't a solution for the original equation, because $(-a)^2 = a^2$ but $-a\neq a$ so you have to check if the solutions are solutions of the original equation too!

And for this reason the unique solution is $x=1$, (note that $+\sqrt{•}$ is always positive and for this reason $+\sqrt{•}\neq -1$)

  • $\begingroup$ Your explanation is confusing at best, probably wrong. The problem lies in the step where he squares both sides - the problem is that you're not forming an equivalence and therefore it's questionable if he's right in all steps. The solution is not $x\in\{-1,1\}$. $\endgroup$ – skyking Apr 6 '17 at 7:52
  • $\begingroup$ Can you read the complete answer please :)? I was modifying it. $\endgroup$ – Rubén Ballester Apr 6 '17 at 7:53
  • $\begingroup$ I'm sorry but I am in a mobile phone and I submitted the answer before I wrote the complete answer $\endgroup$ – Rubén Ballester Apr 6 '17 at 8:02
  • $\begingroup$ It's only small changes since my comment, I still think it's confusing at best. The important step is that for solving the equation the very first step is problematic and what introduces the deviation, it's the nature of this step that influences the whole. $\endgroup$ – skyking Apr 6 '17 at 8:06
  • $\begingroup$ @skyking what part? I want to write a correct answer. $\endgroup$ – Rubén Ballester Apr 6 '17 at 8:06

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